# Function space and abelian group.

Suppose that $$G$$ is a connected Hausdorff topological group. Let consider $$Map([0,1],G)$$ the space of all continuous paths in $$G$$. Clearly $$H=Map([0,1],G)$$ is a topological group (compact-open topology). How to prove that $$H^{ab}=Map([0,1],G^{ab})$$ (as topological groups) where $$H^{ab}$$ is the abelianisation of $$H$$.

• Consider $\mathrm{Map}(K,G)\to\mathrm{Map} (K,G^{ab}),\ f\mapsto \pi\circ f$ where $\pi:G\to G^{ab}$. Its kernel is just $[H,H]$. What is needed to show that the induced map is a homeomorphism? – Berci Dec 22 '19 at 21:46
• @Berci but why $f\mapsto \pi\circ f$ is surjective ? – lab Dec 22 '19 at 21:53
• Yes, good question.. – Berci Dec 22 '19 at 21:55
• @Berci, Just to be sure, are you claiming that is a known fact? I can't distinguish if you are giving a hint or you are asking a question ? :) thank you! – lab Dec 22 '19 at 22:20
• First I intended to give a hint, but your question made me uncertain as well.. – Berci Dec 22 '19 at 23:57

This is indeed the case when $$G$$ is locally compact and the commutator subgroup $$[G,G]$$ is closed in $$G$$ (this is not always the case). I am not sure about the general case. Here is a proof. I will need

Theorem. Let $$G$$ be a Hausdorff locally compact topological group, $$H is a closed subgroup. Then the quotient map $$q: G\to G/H$$ is a Hurewicz-fibration, i.e. it satisfies the homotopy-lifting property. In particular, it satisfies the path-lifting property.

Remark. In general, even if $$G$$ is abelian and metrizable, but is not locally compact, the quotient map $$q$$ need not be a locally trivial fiber bundle.

The only reference for this theorem that I know is Theorem 15 in

E. G. Skljarenko, "On the topological structure of locally bicompact groups and their quotient spaces", Fifteen Papers on Topology and Logic, American Mathematical Society Translations, Series 2 (1964), Volume 39.

The author proves a more general result and credits this theorem to Borel and Serre but I could not find this theorem in their work, so maybe he is being too generous.

In particular, if $$H$$ is a closed normal subgroup of a locally compact group $$G$$, every path $$p: [0,1]\to G/H$$ with $$p(0)=[e]$$ lifts to a path $$\tilde{p}: I=[0,1]\to G$$, $$p(0)=e$$. Applying this to $$H=[G,G]$$ (assuming $$H$$ is closed!), we obtain that $$Map(I, G)\to Map(I, G^{ab})$$ is surjective. The kernel of this homomorphism equals $$Map(I, [G,G])$$, which is the commutator subgroup of $$Map(I, G)$$. Thus, we obtain a natural continuous bijection $$\iota: (Map(I, G))^{ab}\to Map(I, G^{ab}).$$ In order to check continuity of the inverse to $$\iota$$, notice that the Hurewicz fibration property also yields a continuous section $$\sigma$$ for the map $$Map(I, G)\to Map(I, G^{ab})$$ Composing $$\sigma$$ with the quotient map $$Map(I, G)\to (Map(I, G))^{ab}$$ gives a continuous inverse of $$\iota$$. Thus, $$\iota$$ is an isomorphism of topological groups.